From two towns 507km apart, Dave and Bill set out to meet each other. Dave travels 1km the first day, 3 the second, 5 the third, and so on in an arithmetic sequence, while Bill travels 2km the first, 6 the second, 10 the third, etc. How many days after they started will they meet?

you want the total distance traveled by both to be at least 507. So, using your two sequences,

n/2 (2*1 + (n-1)*2) + n/2 (2*2 + (n-1)*2) >= 507

You have two arithmetic sequences.

Dave´s sequence 1 , 3 , 5 ...

the initial term of a1 = 1 and common difference d = 2

nth term of the sequence:

an = a1 + ( n - 1 ) ∙ d

an = 1 + ( n - 1 ) ∙ 2

an = 1 + 2 n - 2

an = 2 n - 1

The sum of the n members of a arithmetic sequence:

Sn = ( n / 2 ) ( a1 + an )

Sn = ( n / 2 ) ( 1 + 2 n - 1 )

Sn = ( n / 2 ) ∙ 2 n

Sn = n²
for Dave

Bill´s sequence 2 , 6 , 10 ...

the initial term of a1 = 2 and common difference d = 4

nth term of the sequence

an = a1 + ( n - 1 ) ∙ d

an = 2 + ( n - 1 ) ∙ 4

an = 2 + 4 n - 4

an = 4 n - 2

The sum of the n members of a arithmetic sequence:

Sn = ( n / 2 ) ( a1 + an )

Sn = ( n / 2 ) ( 2 + 4 n - 2 )

Sn = ( n / 2 ) ∙ 4 n

Sn = 2 n²
for Bill

The combined distance = 507 km so:

n² + 2 n² = 507

3 n² = 507

n² = 507 / 3

n² = 169

n = ± √169

n = ± 13

Number of days can't be negative so:

n = 13

Prooof:

Dave travel:

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 169 km

Bill travel:

2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38 + 42 + 46 + 50 = 338 km

169 km + 338 km = 507 km

My bad. I was using the sequence 2,4,6,8...

rather than 2,6,10,14,...
go with Bosnian.

To find out how many days after they started Dave and Bill will meet, we need to determine when the total distances they have traveled will add up to the distance between the two towns.

Let's start by figuring out how many terms of the arithmetic sequence Dave and Bill need to travel before they meet. We can do this by finding the common difference between their distances.

Dave's distances form an arithmetic sequence with a first term of 1km and a common difference of 2km (1, 3, 5, ...).
Bill's distances also form an arithmetic sequence with a first term of 2km and a common difference of 4km (2, 6, 10, ...).

To find the number of terms, we can set up an equation using the formula for the nth term of an arithmetic sequence:

dave_nth_term = first_term + (n - 1) * common_difference
bill_nth_term = first_term + (n - 1) * common_difference

For Dave:
1 + (n - 1) * 2 = 507

For Bill:
2 + (n - 1) * 4 = 507

Simplifying these equations, we get:
2n - 1 = 507
4n - 2 = 507

Solving these equations, we find:
2n = 508
4n = 509

Therefore, n is approximately 254 for both Dave and Bill.

Since Dave and Bill start on the same day, they will meet after 254 days.